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Quantitative Aptitude Quiz
Quantitative Aptitude Quiz for Railway and SSC covers an important part of the exam and hence it must be prepared thoroughly. SSC, Railways & many other government job exams are scheduled to be held in the upcoming months. It needs complete understanding of the basic concepts along with thorough understanding. This page will provide you all the quizzes of the Arithmetic section of various exams such as RRB NTPC, SSC CHSL, SSC MTS etc. Practice the quantitative aptitude quizzes given below and surpass the high cut off marks in the exam.
ExamsCart is providing free quizzes to help all the aspirants with the SSC & Railway Quantitative Aptitude section by providing links to attempt quizzes. You can attempt quizzes as per the exam interface before the actual exam by attempting the quizzes on the web browser. It’s time to pull up your socks and burn the midnight oil by practicing the SSC Quantitative Aptitude quiz daily.
Quantitative Aptitude Quiz: Set 94
1.If the ratio between the profit and sale price of an article is 1:5, then the ratio between the sale price and the cost price of that article is:
a. 3:2
b. 4:3
c. 5:4
d. 6:5
c. 5:4 Explanation: Profit: Sale Price= 1: 5; => Profit = Sale Price/5; Cost Price = Sale price – Profit; Cost Price = Sale Price– Sale Price/5; => Cost Price = 4* Sale Price/5; Sale Price: Cost Price = 5: 4;
2.What percent of 1 day is 36 minutes?
a. 25%
b. 2.5%
c. 3.6%
d. 0.25%
a. 25% Explanation: 1 day = 24 * 60 minutes = 1440 minutes; The required percentage = 36*100/1440=2.5%.
3. A car covers a certain distance in 25 hrs. If it reduces the speed
by 1/5, the car covers 200 km less in that time. The speed of
car is
a. 60 km/hr
b. 30 km/hr
c. 40 km/hr
d. 50 km/hr
c. 40 km/hr Explanation: Let the distance be x km and speed be y kmph; x/y = 25 => x =25y; ——–eq.(i.) (x-200)/(4y/5) = 25 =>x-200 = 20y; —- eq.(ii.) Put value of x from eq.(i) to eq.(ii.)- 25y-200 = 20y; 5y = 200 => y=40 kmph;
4.If x2+y2+z2 = 14 and xy + yz + zx = 11, then the value of (x+y+z)2 is
a. 16
b. 25
c. 36
d. 49
c. 36 Explanation: x2+y2+z2 = 14 ———–eq.(i.) xy + yz + zx = 11 ————-eq.(ii.) Adding eq.(i.) and 2 x eq.(ii.), we get- x2+y2+z2 +2(xy + yz + zx) = 14 + 2*11=36; (x+y+z)2 =36;
5.A positive number when decreased by 4, is equal to 21 times the reciprocal of this number. The number is:
a. 3
b. 7
c. 5
d. 9
b. 7 Explanation: Suppose the number =x; x-4 = 21*(1/x); => x2-4x-21=0; x2-(7-3)x-21=0; x2-7x+3x-21=0; => x(x-7)+3(x-7)=0; (x-7)(x+3)=0; => x=7, -3;
6.
a. 48
b. 1792
c. 4096
d. 570
c. 4096 Explanation: x/y = 12/4 =3; Hence, (x+y)x/y=(12+4)3 =163 =4096;
7.In a triangle PQR, PQ = PR and ∠Q is twice that of ∠P. Then ∠Q is equal to
a. 72°
b. 36°
c. 144°
d. 108°
a. 72°
8.If the length of a chord of a circle is 16 cms. and is at a distance of 15 cm from the centre of the circle, then the radius of the circle (in cm) is:
a. 15
b. 16
c. 17
d. 34
c. 17
9.On 24th May 2008 the maximum temperature of Delhi, Kolkata and Mumbai were recorded as 35°C, 33°C and 34°C respectively. What was the maximum temperature of Chennai so that the average maximum temperature of those cities would be 35°?
a. 34° C
b. 35° C
c. 36° C
d. 38° C
d. 38° C Explanation: Average temperature of all cities including Chennai = 35 + 33 + 34 + x= 102 + x; (102 + x )/4 = 35; x = 140 -102 = 38;
10.The diameter of a sphere is twice the diameter of another sphere. The curved surface area of the first and the volume of the second are numerically equal. The numerical value of the radius of the first sphere is
a. 3
b. 24
c. 8
d. 16
b. 24 Explanation: Let the diameter of first sphere is d and diameter of second sphere is D. d=2D; pi * d2 = 1/3 * pi * (D)3; put value of d in the above equation- 4D2 = 1/3 *(D3); => D = 12 cms; d=2*12 = 24 cms.
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